Question: $f(x)=\dfrac{1}{1-2x}$ We know that $f(x)=1+2x+{{(2x)}^{2}}+{{(2x)}^{3}}+...+{{(2x)}^{n}}+...$ for $x\in \left(-\dfrac12,\dfrac12\right)$. Using this fact, find the power series for $g(x)=\dfrac{2}{(1-2x)^2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1+2x+{{(2x)}^{2}}+...$ (Choice B) B $2+8x+24{{x}^{2}}+...$ (Choice C) C $2x+{{(2x)}^{2}}+{{\left( {{(2x)}^{2}} \right)}^{3}}+...$ (Choice D) D $1+2x+3{{x}^{2}}+...$ (Choice E) E $2-8x+24{{x}^{2}}+...$
Explanation: First note that $~g(x)=f\,^\prime(x)\,$. Hence, to find the series represented by $~g(x)\,$, differentiate $~f(x)$ term-by term $ f(x)=1+2x+4x^2+8x^3+16x^4+...+2^n x^n+...$ $ g(x)=f\,^\prime(x)=2+8x+24{{x}^{2}}+64x^3+...+n\cdot 2^n x^{n-1}+...$